Suppose that $ 1 < p < \infty $ and that $T$ is a linear operator from $L_p[0, 1]$ into itself with the property that if $(f_n)$ is a sequence in $L_p[0, 1]$ that converges almost everywhere to some $ f \in L_p[0, 1]$, then $(Tf_n)$ converges almost everywhere to $T_f$. Prove that $T$ is bounded.
Attempted proof
This seems to follow immediately from the Uniform Boundedness Principle, which states:
Let $F$ be a nonempty family of bounded linear operators from a Banach space $X$ into a normed space $Y$. If $\sup \{ \lVert Tx \rVert : T \in F \}$ is finite for each $x$ in $X$, then $ \sup \{ \lVert T \rVert : T \in F \} $ is finite.
In our problem, we merely need to consider $L_p[0, 1]$ as the Banach space $X$ and the result follows. $\quad \square$
It's either this simple, or I'm missing a point.
Closed Graph Theorem is closer. But we need to fulfil the premises of Closed Graph Theorem:
We need to prove that, if $(f_{n})\subseteq L^{p}[0,1]$ converges in $L^{p}[0,1]$ to $f$ and $(T(f_{n}))$ converges to $g$ in $L^{p}[0,1]$, then $g=T(f)$.
Note that $L^{p}[0,1]$ convergence implies a pointwise a.e. convergence of a subsequence. So, we have, say, $f_{n_{k}}\rightarrow f$ a.e. for some subsequence $(f_{n_{k}})$. Then by assumption we know that $T(f_{n_{k}})\rightarrow T(f)$ a.e.
For this subsequence $(T(f_{n_{k}}))$, it still holds true that $T(f_{n_{k}})\rightarrow g$ in $L^{p}[0,1]$. So, we have, say, $T(f_{n_{k_{l}}})\rightarrow g$ a.e. for some subsequence $(T(f_{n_{k_{l}}}))$. Since it still holds true that $T(f_{n_{k_{l}}})\rightarrow T(f)$ a.e., we conclude that $g=T(f)$ a.e.
Under the a.e. identification in $L^{p}[0,1]$, then $g=T(f)$, the result follows.